Science:Math Exam Resources/Courses/MATH104/December 2016/Question 13

MATH104 December 2016

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Question 13

A bucket is $60{\text{cm}}$ high, has a radius of $40{\text{cm}}$ at the top and $10{\text{cm}}$ at the bottom. Water is being dumped into the bucket at a rate of $1{\text{L}}/{\text{min}}$. How fast is the water level rising when the water is 30 cm deep? You may use the fact that the volume of a trunctated cone with height ${\textstyle h}$, radius at the bottom ${\textstyle r_{1}}$ and radius at the top ${\textstyle r_{2}}$, is given by ${\textstyle V={\frac {1}{3}}\pi h(r_{2}^{2}+r_{1}r_{2}+r_{1}^{2})}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Let ${\textstyle h}$ be the height of the water and ${\textstyle r}$ the corresponding radius. Find their relation.

Hint 2
Use chain rule to get ${\frac {dh}{dt}}{\bigg |}_{h=30}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let ${\textstyle h}$ be the height of the water and ${\textstyle r}$ the corresponding radius. Then the volume would be given by: $$V={\frac {1}{3}}\pi h(r^{2}+10r+100).$$ Since we know that ${\frac {dV}{dt}}=1000{\text{cm}}^{3}/{\text{min}}$, we’ll need to replace ${\textstyle h}$ or ${\textstyle r}$. Let’s find an equation relating them: From which we see that $$h=2r-20.$$ Thus, our equation becomes: $$V={\frac {1}{3}}\pi h(r^{2}+10r+100)={\frac {\pi }{3}}(2r-20)(r^{2}+10r+100)={\frac {\pi }{3}}(2r^{3}-2000)$$ Then, applying the chain rule, we have $${\frac {dV}{dt}}={\frac {\pi }{3}}\cdot 6r^{2}{\frac {dr}{dt}},$$ and then we plug ${\frac {dV}{dt}}=1000$ and solve for ${\frac {dr}{dt}}$ to get $$1000=2\pi r^{2}{\frac {dr}{dt}}\implies {\frac {dr}{dt}}{\bigg |}_{r=25,h=30}={\frac {500}{25^{2}\cdot \pi }}={\frac {4}{5\pi }}$$ Here, $r=25$ when $h=30$ follows from the relation $h=2r-20.$. Since ${\textstyle h=2r-20}$, we finally get $${\frac {dh}{dt}}{\bigg |}_{r=25,h=30}=2{\frac {dr}{dt}}{\bigg |}_{r=25,h=30}={\frac {8}{5\pi }}$$ Answer: $\color {blue}{\frac {8}{5\pi }}{\text{cm}}/{\text{min}}$